Question

1. A study was conducted to examine whether the perception of service quality at hotels differed by gender. Hotel guests were randomly selected to rate hotel service, and each rating was then transferred into a standardized score. By comparing the scores observed from the sample of male guests and the sample of female guests, researchers then wished to determine if there is evidence of a difference in average rating when awarded by males, μ ₁, and the average rating when awarded by females, μ ₂. A summary of the sampled guest scores are provided in the table below :

Group Statistics
	GROUP	n	Mean	Std. Deviation
Quality Score	males females	8 22	5.55 4.26	7.41 5.80

2. 
   i) Carry out a hypothesis test for a significant difference between the two population means, at significance level α = 0.05.
   The hypotheses being tested are:
   H ₀: μ ₁ - μ ₂ = 0
   H _a: μ ₁ - μ ₂ ≠ 0.
   Complete the test by filling in the blanks in the following:
   • An estimate of the difference in population means is .
   • Assuming the population standard deviations are equal, σ₁=σ₂, the pooled estimate, s_p, is  and the standard error is .
   • The distribution is  (examples: normal / t12 / chisquare4 / F5,6).

The test statistic has value TS=  .
Testing at significance level α = 0.05, the rejection region is:
less than  and greater than  (3 dec places).
There  (is evidence/is no evidence) to reject the null hypothesis, H ₀ of no difference between the two population means, μ ₁ and μ ₂.

ii) Estimate the difference in population means by calculating a 95% confidence interval.
The difference between the population means, the mean of population 1, μ ₁, minus the mean of population 2, μ ₂, is estimated to be between________  and__________ .

Answer 1

(i) H₀: μ₁ - μ ₂ = 0
H _a: μ ₁ - μ ₂ ≠ 0.

An estimate of the difference in population means is = 5.55 - 4.26 = 1.29

Assuming the population standard deviations are equal, σ₁=σ₂, the pooled estimate, s_p, is  and the standard error is .

Pooled estimate= s_p = sqrt [{(n₁ -1)s₁² + (n₂ -1)s₂²}/(n₁ + n₂- 2)]

= sqrt [(7 * 7.41 * 7.41 + 21 * 5.80 * 5.80)/(8 + 22 - 2)]

= 6.2416

Standard error of proportion = se_p = sqrt [s_p² * (1/n₁ + 1/n₂)] = sqrt [6.2416² * (1/8 + 1/22)] = 1.0315

The distribution is t test statistic.

Degree of freedom = 8 + 22 - 2 = 28

Test statistic

t = (M₁-M₂) / se₀ = (5.55 - 4.26)/1.0315 = 1.251

Here Testing at significance level α = 0.05, the rejection region is

t_critical = TINV(0.05, dF = 28) = 2.0484

so here rejection region is where t > 2.0484 and t < -2.0484

There is not enough evidence to reject the null hypothesis.

(ii) 95% confidence interval = (M₁ -M₂) +- t_critical se

= (5.55 - 4.26) +- 2.0484 * 1.0315

= (-0.823, 3.403)